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QveST [7]
3 years ago
10

(5x-6) squared please answer​

Mathematics
2 answers:
Aleonysh [2.5K]3 years ago
7 0
Check photo for answer! i have included work

pashok25 [27]3 years ago
6 0

Answer:

25 x 2 − 60 x + 36

Step-by-step explanation:

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A farmer has a feed trough in the shape of rectangular prism with a height of 40 cm and a base that is 30 cm wide by 200 cm long
inysia [295]

Answer:

Step-by-step explanation:

You're welcome :)

5 0
3 years ago
An internet charges a flat rate of $6.50 for shipping. If x represents the cost of items purchased and y represents the total co
Oxana [17]

<em><u>Question:</u></em>

An Internet business charges a flat rate of $6.50 for shipping. If x represents the cost of the items purchased and y represents the total cost with shipping, which equation best represents the relationship between x and y?

<em><u>Answer:</u></em>

y = 6.5 + x represents the relationship between x and y

<em><u>Solution:</u></em>

Given that,

An internet charges a flat rate of $6.50 for shipping

Let "x" be the cost of items purchased

Let "y" be the total cost with shipping

Therefore,

Total cost with shipping = flat rate of 6.50 + cost of items purchased

Thus, we get,

y = 6.50 + x

y = 6.5 + x

Thus the above equation represents the relationship between x and y

8 0
3 years ago
a rubber ball is dropped on a hard surface takes a sequence of bounces, each one 3/5 as high as the preceeding one. If this ball
alex41 [277]

Answer:

36.172

Step-by-step explanation:

10+6+6+3.6+3.6+2.16+2.16+1.296+1.296=36.172

8 0
4 years ago
6(5x-3) what is the answer to this
Zarrin [17]

Answer:

= 30x − 18

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
20 POINTS!! ASAP, PLS SHOW WORK TYY
Sergeeva-Olga [200]

Answer:

\sin(\theta)=-\sqrt5/5\text{ and } \csc(\theta)=-\sqrt5\\\cos(\theta)=2\sqrt5/5\text{ and } \sec(\theta)=\sqrt5/2\\\tan(\theta)=-1/2\text{ and } \cot(\theta)=-2

Step-by-step explanation:

First, let's determine which quadrant our angle θ lies in.

Remember ASTC, where:

Everything is positive in QI,

Only sine (and cosecant) is positive in QII,

Only tangent (and cotangent) is positive in QIII,

And only cosine (and secant) is positive in QIV.

Since our tangent is negative, and our cosine is positive, this means that our θ <em>must</em> be in QIV.

In QIV, sine is negative, tangent is negative, and cosine is positive.

With that, let's figure out the remaining trig ratios.

We know that:

\tan(\theta)=-1/2

Remember that tangent is the ratio of the opposite side to the adjacent side.

Let's figure out our hypotenuse using the Pythagorean Theorem:

a^2+b^2=c^2

Substitute 1 for a and 2 for b (we can ignore the negative since we're squaring anyways). This yields:

(1)^2+(2)^2=c^2

Square:

1+4=c^2

Add:

c^2=5

Take the square root:

c=\sqrt{5}

So, our square root is √5.

So, our three sides are: Opposite=1, Adjacent=2, and Hypotenuse=√5.

Sine and Cosecant:

Remember that:

\sin(\theta)=opp/hyp

Substitute 1 for the opposite and √5 for the hypotenuse. This yields:

\sin(\theta)=1/\sqrt5

Rationalize:

\sin(\theta)=\sqrt5/5

And since our angle is in QIV, we add a negative:

\sin(\theta)=-\sqrt5/5

Cosecant is simply the reciprocal of sine. So:

\csc(\theta)=-\sqrt5

Cosine and Secant:

Remember that:

\cos(\theta)=adj/hyp

Substitute 2 for the adjacent and √5 for the hypotenuse. This yields:

\cos(\theta)=2/\sqrt5

Rationalize:

\cos(\theta)=2\sqrt5/5

Since our angle is in QIV, cosine stays positive.

Secant is the reciprocal of cosine. So:

\sec(\theta)=\sqrt5/2

Tangent and Cotangent:

We were given that:

\tan(\theta)=-1/2

To find cotangent, flip:

\cot(\theta)=-2

And we're done!

7 0
3 years ago
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